Long-Term Care Across Multiple Care Levels

Why Model More Than One Care Level?

A long-term care contract may pay different amounts depending on the insured’s care needs. An insured may move from healthy status into a lower care level, later require a higher level of care, or recover to a less intensive state. Death ends future care payments. A model with only one “in care” state cannot distinguish these paths or assign different benefit rates to different care needs.

The central task is to translate the product description into states and permitted transitions before valuing benefits. Once the model is stated, the expected present value of continuous care payments can be found from a system of linear equations. The equations account for both the benefit paid while occupying each state and the possibility of moving to another state.

Definition: A state is a category that describes the insured’s current condition for purposes of the model. In this tutorial, $H$ denotes healthy status, $C_1$ a lower care level, $C_2$ a higher care level, and $D$ death. A transition intensity $q_{ij}$ is the instantaneous rate of transition from state $i$ to a different state $j$, conditional on occupying state $i$. Death is absorbing: once the insured enters $D$, no further care benefit is payable.

The model must state which transitions are possible. For example, a contract may allow recovery from $C_1$ to $H$, an upgrade from $C_1$ to $C_2$, and a move from $C_2$ back to $C_1$. If a transition is not allowed by the model, its intensity is zero. The rates may be constant over time in a particular calculation, but that is an assumption, not a property of every long-term care model.

Generator and State-Value Equations

Use a row-generator convention: each off-diagonal entry is the intensity of the transition from the row state to the column state, and each diagonal entry is the negative sum of all intensities leaving that state. The transient states are those other than death. When forming the transient generator, retain the full exit rate on each diagonal, including the rate of death, even though the death column is omitted.

Definition: Let $Q_T$ be the generator restricted to the transient states, with death omitted from its columns but included in each relevant diagonal exit rate. Let $b_i$ be the continuous benefit payment rate while the insured occupies transient state $i$, and let $w_i$ be the expected present value of future benefits starting in state $i$. Under constant transition intensities, a constant force of interest $\delta>0$, no transition lump sums, and benefits paid continuously while in each state, $$ (\delta I-Q_T)\boldsymbol{w}=\boldsymbol{b}. $$ The value on entering death is zero.

The benefit rate $b_i$ is a cash flow per unit time, not a lump sum paid upon entering state $i$. A healthy-state benefit can be zero while care-state benefits are positive. In the system above, a transition affects value through the chance of leaving the current state and the value available in the destination state. This formulation does not include benefits payable specifically on transitions; those would require additional terms.

Why the Valuation Equations Hold

Consider a short interval of length $dt$ starting in transient state $i$. Write $\lambda_i$ for the total intensity of all exits from $i$, including death. To first order in $dt$, the probability of no transition is $1-\lambda_i dt$, and the probability of a transition to a particular transient state $j$ is $q_{ij}dt$. The benefit paid during the interval is $b_i dt$ to first order.

Discounting over this short interval gives the first-step relation

w_i=b_i\,dt+e^{-\delta dt}\left[(1-\lambda_i dt)w_i+\sum_{j\ne i}q_{ij}dt\,w_j\right]+o(dt).

The sum includes transient destination states; a transition to death contributes zero continuation value. Use $e^{-\delta dt}=1-\delta dt+o(dt)$, subtract $w_i$ from both sides, divide by $dt$, and let $dt$ tend to zero. This yields

\delta w_i=b_i-\lambda_i w_i+\sum_{j\ne i}q_{ij}w_j.

Since the diagonal of $Q_T$ is $-\lambda_i$, this equation for every transient state is exactly $(\delta I-Q_T)\boldsymbol{w}=\boldsymbol{b}$. This argument also explains why every exit, including death, belongs in the diagonal rate.

Worked Example: Valuing Two Care Levels with Recovery

A hypothetical contract has states $H,C_1,C_2,D$. From $H$, the insured enters $C_1$ at intensity $0.10$ per year or dies at intensity $0.10$. From $C_1$, recovery to $H$, an upgrade to $C_2$, and death each have intensity $0.10$. From $C_2$, a move back to $C_1$ and death each have intensity $0.10$. The continuous benefit rates are \$0, \$20,000, and \$50,000 per year in $H,C_1,C_2$, respectively. The force of interest is $\delta=0.10$. Find the values from each starting state.

Method. Form the transient generator using the stated exits, then solve the state-value equations. Work in thousands of dollars so the benefit vector is $(0,20,50)^\mathsf{T}$.

The transient generator, with states ordered $H,C_1,C_2$, is

Q_T= \begin{pmatrix} -0.20&0.10&0\\ 0.10&-0.30&0.10\\ 0&0.10&-0.20 \end{pmatrix}.

For example, the $C_1$ diagonal is $-0.30$, because the total exit intensity is $0.10+0.10+0.10=0.30$. Adding $\delta I$ gives

\begin{pmatrix} 0.30&-0.10&0\\ -0.10&0.40&-0.10\\ 0&-0.10&0.30 \end{pmatrix} \begin{pmatrix}w_H\\w_1\\w_2\end{pmatrix} = \begin{pmatrix}0\\20\\50\end{pmatrix}.

The three equations are $0.30w_H-0.10w_1=0$, $-0.10w_H+0.40w_1-0.10w_2=20$, and $-0.10w_1+0.30w_2=50$. Multiply each equation by 10:

3w_H-w_1=0,\qquad -w_H+4w_1-w_2=200,\qquad -w_1+3w_2=500.

The first equation gives $w_1=3w_H$. The third gives $w_2=(500+w_1)/3$. Substitute both into the second: $ -w_H+4(3w_H)-(500+3w_H)/3=200$. Thus $10w_H=1100/3$, so $w_H=110/3$. It follows that $w_1=110$ and $w_2=610/3$, all in thousands of dollars.

Therefore, to the nearest cent, the values are $w_H=\$36,666.67$, $w_1=\$110,000.00$, and $w_2=\$203,333.33$. Checking the care equations in thousands gives $-0.1(36.6667)+0.4(110)-0.1(203.3333)=20$ and $-0.1(110)+0.3(203.3333)=50$, as required. This method avoids the mistake of treating $C_2$ as a terminal state: it can return to $C_1$, so its continuation value matters.

Starting-State Mixtures and Payment Units

If the initial state is uncertain, the expected present value is the probability-weighted average of the state values. The weighting probabilities describe the state at time zero; they are not transition probabilities over a later period. Calculate each state value first, then apply the initial-state probabilities.

Worked Example: An Initial Mixture Across Three States

A hypothetical model has states $H,L,S,D$, where $L$ is a lower care level and $S$ is a more intensive care level. From $H$, the insured moves to $L$ at intensity $0.20$ or dies at $0.10$. From $L$, the intensities to $H,S,D$ are each $0.10$. From $S$, the intensities to $H,L,D$ are each $0.10$. Benefits are \$0, \$30,000, and \$60,000 per year in $H,L,S$; $\delta=0.10$. At issue, the probabilities of $H,L,S$ are $0.50,0.30,0.20$. Find the expected present value at issue.

Method. Solve the state values in thousands of dollars, then weight them by the initial-state probabilities. The transient generator has diagonal entries $-0.30$ in all three states: each total exit rate is $0.30$.

(\delta I-Q_T)\boldsymbol{w}=\boldsymbol{b} \quad\Longrightarrow\quad \begin{cases} 0.40w_H-0.20w_L=0,\\ -0.10w_H+0.40w_L-0.10w_S=30,\\ -0.10w_H-0.10w_L+0.40w_S=60. \end{cases}

Multiplying by 10 gives $4w_H-2w_L=0$, $-w_H+4w_L-w_S=300$, and $-w_H-w_L+4w_S=600$. The first equation gives $w_H=w_L/2$. From the second, $w_S=4w_L-w_H-300=3.5w_L-300$. Substitute into the third: $-0.5w_L-w_L+4(3.5w_L-300)=600$. Hence $12.5w_L=1800$, so $w_L=144$, $w_H=72$, and $w_S=204$, in thousands of dollars.

The initial value is $0.50(72)+0.30(144)+0.20(204)=36+43.2+40.8=120$ thousand dollars, or \$120,000.00. As a check, substitution into the three original equations gives benefit rates $0,30,60$, respectively. This calculation avoids assigning the value from one presumed starting state when the problem specifies an initial mixture.

Worked Example: Converting Monthly Benefits to Annual Rates

A hypothetical contract uses the $H,C_1,C_2,D$ transition intensities from the first worked example, with $\delta=0.10$. It pays \$2,000 per month in $C_1$ and \$5,000 per month in $C_2$, continuously while the insured occupies each state. No benefit is paid in $H$. Find the value from $H$.

Method. The transition intensities and valuation equations use years as the time unit, so convert each monthly amount to a continuous annual payment rate: $12(\$2,000)=\$24,000$ and $12(\$5,000)=\$60,000$. In thousands, the benefit vector is $(0,24,60)^\mathsf{T}$.

The equations are $0.30w_H-0.10w_1=0$, $-0.10w_H+0.40w_1-0.10w_2=24$, and $-0.10w_1+0.30w_2=60$. Multiplying by 10 gives $3w_H-w_1=0$, $-w_H+4w_1-w_2=240$, and $-w_1+3w_2=600$. The first equation gives $w_1=3w_H$, and the third gives $w_2=(600+w_1)/3$. Substitution into the middle equation yields $10w_H=1320/3$, so $w_H=44$, $w_1=132$, and $w_2=244$, all in thousands.

The value from healthy status is therefore \$44,000.00. Checking the equations gives $0.3(44)-0.1(132)=0$, $-0.1(44)+0.4(132)-0.1(244)=24$, and $-0.1(132)+0.3(244)=60$. This method avoids using monthly benefit amounts directly with annual transition intensities and an annual force of interest.

Common Traps and Model Checks

The state-value equations are compact, but they depend on a correctly specified model. In particular, do not omit an exit from a diagonal just because it leads to death or because it does not lead to another care state. Do not put a continuous care benefit into a transition term: it is a payment rate while occupying a state. Finally, use a consistent time unit for intensities, interest, and benefit rates.

A useful check on the diagonal and the valuation method is a two-state special case. Suppose $H$ moves to care $C$ and to death at intensities $0.10$ each; from $C$, recovery to $H$ and death also have intensities $0.10$ each. Let $\delta=0.10$ and let the care benefit be \$40,000 per year. In thousands, the correct equations are $0.20w_H-0.10w_C=0$ and $-0.10w_H+0.30w_C=40$. The first gives $w_H=0.5w_C$; substituting into the second gives $0.25w_C=40$. Thus $w_C=160$ and $w_H=80$, or \$160,000.00 and \$80,000.00. If the initial probabilities are $0.80$ in $H$ and $0.20$ in $C$, the mixture is $0.80(80{,}000)+0.20(160{,}000)=\$96,000.00$.

The methods here—constructing linked states and valuing state-contingent cash flows—are aligned with the SOA Exam ALTAM (Advanced Long-Term Actuarial Mathematics) syllabus.

Key takeaway: Define the care states and permitted transitions first. Include every exit intensity, including death, in the appropriate generator diagonal; then solve $(\delta I-Q_T)\boldsymbol{w}=\boldsymbol{b}$ using benefit rates and time units consistent with the model.

Check Your Understanding

For each problem, state the relevant model assumptions and keep the time units consistent.

A model has healthy state $H$, care state $C$, and death $D$. The intensities from $H$ to $C$ and $D$ are $0.15$ and $0.05$; from $C$ to $H$ and $D$ they are $0.10$ and $0.10$. If $\delta=0.10$ and the care benefit is \$30,000 per year, find the value from $H$. (Answer: \$45,000.00)
In a three-state model $H,L,S$ before death, the only exits from $H$ are $H\to L$ at $0.20$ and $H\to D$ at $0.10$. What is the diagonal entry for $H$ in the transient generator, and why? (Answer: $-0.30$ per year; both exits count.)
Using the model in Question 2, suppose $\delta=0.10$, $L$ has only $L\to H$ and $L\to D$ exits, each at $0.10$, and the benefits are \$0 in $H$ and \$20,000 per year in $L$. Find the value from $H$. (Answer: \$20,000.00)
A model has the same transition intensities as the first worked example, $\delta=0.10$, and continuous benefits of \$1,500 per month in $C_1$ and \$3,000 per month in $C_2$. Find the value from $H$. (Answer: \$26,500.00)
In the second worked example, change the initial probabilities to $0.20$ in $H$, $0.50$ in $L$, and $0.30$ in $S$. Find the expected present value at issue. (Answer: \$147,600.00)

What You'll Learn